Electron launched at an angle in a parallel plate field.

[Esoremada]

Homework Statement

http://puu.sh/6i6HY.png

Homework Equations

Fe = q*Ee

The Attempt at a Solution

a)
d_x = v_x*t
0.0682 = 4.92*10⁶ * cos(68) * t
t = 0.0682 / [4.92*10⁶ * cos(68)]
= 3.70036*10^-8 seconds

d_y = d_yi + v_y*t + at²
0 = 0 + 4.92*10⁶ * [0.0682 / [4.92*10⁶ * cos(68)]] * sin(68) + a * [0.0682 / [4.92*10⁶ * cos(68)]]²
a = -4.92*10⁶ * [0.0682 / [4.92*10⁶ * cos(68)]] * sin(68) / [0.0682 / [4.92*10⁶ * cos(68)]]²
= -1.23278 * 10¹⁴ m/s²

mass of electron = 9.10938291 * 10^-31 kg
charge of electron = 1.60217657 * 10^-19 C

Fe = ma
= 9.10938291 * 10^-31 * [ -4.92*10⁶ * [0.0682 / [4.92*10⁶ * cos(68)]] * sin(68) / [0.0682 / [4.92*10⁶ * cos(68)]]² ]

Ee * q = Fe
Ee * 1.60217657 * 10^-19 = 9.10938291 * 10^-31 * [ -4.92*10⁶ * [0.0682 / [4.92*10⁶ * cos(68)]] * sin(68) / [0.0682 / [4.92*10⁶ * cos(68)]]² ]
Ee = 9.10938291 * 10^-31 * [ -4.92*10⁶ * [0.0682 / [4.92*10⁶ * cos(68)]] * sin(68) / [0.0682 / [4.92*10⁶ * cos(68)]]² ] / (1.60217657 * 10^-19)I got part b correct, so I must have the correct acceleration - now I can't figure out where I'm going wrong.

 

[Simon Bridge]

Check your reasoning ...
Also: Remember your ballistics?

If you have the acceleration right, then E=a(m/q)
If that gives the wrong E then (a) you have used a bad value for q/m for an electron - maybe a rounding error? or (b) you didn;t get the right acceleration and the correctness of the answer to part b is a fluke.

Best practice is to do all the algebra before putting the numbers in.

 

[ehild]

Homework Statement

http://puu.sh/6i6HY.png

Homework Equations

Fe = q*Ee

The Attempt at a Solution

a)

d = vt
0.0682 = 4.92*10^6 * cos(68) * t
t = 0.0682 / [4.92*10^6 * cos(68)]
= 3.70036*10^-8

The time of flight is correct if you meant in in seconds.

d = di + vt + at^2
0 = 0 + 4.92*10^6 * 3.70036*10^-8 * sin(68) + a * [3.70036*10^-8]^2
a = -4.92*10^6 * 3.70036*10^-8 * sin(68) / [3.70036*10^-8]^2
= -1.23278 * 10^14 m/s^2

Is it the same d as before, when you wrote that d=vt?


The electric field is normal to the capacitor plates. What is the direction of the acceleration? Does the electron accelerate horizontally?

ehild

 

[Esoremada]

Is it the same d as before, when you wrote that d=vt?


The electric field is normal to the capacitor plates. What is the direction of the acceleration? Does the electron accelerate horizontally?

ehild

No I used horizontal velocity and distance to find time, and then used vertical velocity, distance and time to find acceleration. I'll add subscripts to the OP to make that more clear

Check your reasoning ...
Also: Remember your ballistics?

If you have the acceleration right, then E=a(m/q)
If that gives the wrong E then (a) you have used a bad value for q/m for an electron - maybe a rounding error? or (b) you didn;t get the right acceleration and the correctness of the answer to part b is a fluke.

Best practice is to do all the algebra before putting the numbers in.

I took the electron values to many more decimals than the homework system checks (3).

I changed all my calculations to not use any previously calculated values and reformatted the exponents. Do you see any specific incorrect logic or calculation now?

 

[ehild]

The equation

d_y = d_yi + v_y*t + at²

is wrong. Check.

ehild

 

[Esoremada]

The equation

d_y = d_yi + v_y*t + at²

is wrong. Check.

ehild

Oh man, messed up that equation after using it hundreds of times. Thank you very much, that fixed it.

(and I forget the 1/2 both times and the mistakes canceled out which is why I got the right answer for part b )